Now compare back to the original expressions at the top of the answer, and you should convince yourself that this holds true.

WHY IS ENTROPY EQUAL TO REVERSIBLE HEAT FLOW?

Well, let's put it this way. If you do stuff in such a way that you go from an initial state back to a final state equal to the initial state, then you've performed a cyclic process (such as the Carnot cycle).

We denote this as ##ointdS = 0##.

This makes sense, because we know that entropy is a state function, so if ##S_i = S_f##, then

##color(blue)(ointdS = DeltaS = 0).##

But if you perform an ##x## process, and then perform a ##y## process instead of a ##-x## process (where ##x ne y##), then you've performed an irreversible process, which is not necessarily cyclic...

That's not kosher, because ##S_i ne S_f##, and thus, you've performed a path function process! That is, the path apparently does matter, when it shouldn't for a state function.

Hence, it couldn't be that ##DeltaS = (q_"irr")/T##.

If you perform an ##x## process and then a ##-x## process, you've performed a reversible cyclic process. Now, you CAN say this: